Rút gọn :
\(\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2\) phần 2.(-1)5 + ( \(\dfrac{3}{4}\) )2 - \(\dfrac{3}{8}\)
rút gọn
g, \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right).\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\) h,\(\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\dfrac{1}{3}}\right).\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\dfrac{1}{5}}\right)\)
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
Câu 1 : Rút gọn
\(G=\dfrac{6!}{\left(m-2\right)\left(m-3\right)}.\left[\dfrac{\left(m+1\right)!}{5!.\left(m-4\right)!.\left(m+1\right)}-\dfrac{m!}{12.3!.\left(m-4\right)!}\right]\)
Câu 2 : CMR
\(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}< 3\forall n\in N\)
Rút gọn các biểu thức sau:
a. \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}\) - \(\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
b.\(\dfrac{1}{4-3\sqrt{2}}\) - \(\dfrac{1}{4+3\sqrt{2}}\)
c.\(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right)\): \(\sqrt{28}\)
d.\(\dfrac{3}{\sqrt{6}-\sqrt{3}}\)+\(\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
Rút gọn biểu thức trên
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
\(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-\dfrac{1}{8}-\dfrac{27}{16}}{-\dfrac{29}{16}}\)
\(=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}=1\)
Chúc bạn học tốt!!!
P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
Bài 8:
b)\(\left(\dfrac{-4}{3}\right)+\left(\dfrac{-2}{5}\right)+\left(\dfrac{-3}{2}\right)\)
c) \(\dfrac{4}{5}-\left(\dfrac{-2}{7}\right)-\dfrac{-7}{10}\)
d) \(\dfrac{2}{3}-\left[\left(\dfrac{-7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
RÚT GỌN BIỂU THỨC
A= \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\)\(\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
B= \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\)\(\left(\sqrt{6}+11\right)\)
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
Rút gọn biểu thức
a) A=\(2\sqrt{\left(2-\sqrt{5}\right)^2}-\dfrac{8}{3-\sqrt{5}}\)
b) B= \(\left(\dfrac{2\sqrt{x}}{x-4}-\dfrac{1}{\sqrt{x}+2}\right):\left(1+\dfrac{2}{\sqrt{x}-2}\right)\) Với x>0, x khác 4
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
Rút gọn:
\(A=\dfrac{6!}{\left(m-2\right)\left(m-3\right)}.\left[\dfrac{1}{\left(m+1\right)\left(m-4\right)}.\dfrac{\left(m+1\right)!}{\left(m-5\right)!5!}-\dfrac{m\left(m-1\right)!}{12.\left(m-4\right)!3!}\right]\) với \(m\ge5\)
Lời giải:
\(A=\frac{6!}{(m-2)(m-3)}\left[\frac{m!}{(m-4)!.5!}-\frac{m!}{(m-4)!3.4!}\right]\)
\(=\frac{6!}{(m-2)(m-3)}.\frac{m!}{(m-4)!}(\frac{1}{5!}-\frac{1}{3.4!})=\frac{-4}{(m-2)(m-3)}.\frac{m!}{(m-4)!}\)
\(=\frac{-4}{(m-2)(m-3)}.(m-3)(m-2)(m-1)m=-4m(m-1)\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`